Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 17, K = 10;
int sz[N], ch[N], pos[N], p[N], in[N], timer = 0;
vector<int> g[N] = {{0}}, z[N];
int k = K;
vector<int> operator+(const vector<int> &a, const vector<int> &b) {
  vector<int> v(int((a).size()) + int((b).size()));
  merge(begin(a), end(a), begin(b), end(b), begin(v));
  if (k < int((v).size())) v.resize(k);
  return v;
}
int lo, hi;
struct T {
  int n;
  vector<int> nodes;
  vector<vector<int>> t;
  void init() {
    n = int((nodes).size());
    int sz = 1;
    while (sz < n) sz <<= 1;
    t.resize(2 * sz);
    build(1, 0, n);
  }
  void build(int v, int l, int r) {
    if (r - l == 1) return swap(t[v], z[nodes[l]]);
    int m = l + r >> 1;
    build(v << 1, l, m);
    build(v << 1 | 1, m, r);
    t[v] = t[v << 1] + t[v << 1 | 1];
  }
  vector<int> query(int v, int l, int r) {
    if (r <= lo || hi <= l) return {};
    if (lo <= l && r <= hi) return t[v];
    int m = l + r >> 1;
    return query(v << 1, l, m) + query(v << 1 | 1, m, r);
  }
  vector<int> query(int l, int r) {
    if (l > r) swap(l, r);
    lo = l;
    hi = r + 1;
    return query(1, 0, n);
  }
};
vector<T> t;
void dfs(int u, int a) {
  in[u] = ++timer;
  sz[u] = 1;
  p[u] = a;
  g[u].erase(find(begin(g[u]), end(g[u]), a));
  if (g[u].empty())
    ch[u] = int((t).size()), t.emplace_back();
  else {
    int mx = g[u].front();
    for (auto &v : g[u]) {
      if (v == a) continue;
      dfs(v, u);
      sz[u] += sz[v];
      if (sz[mx] < sz[v]) mx = v;
    }
    ch[u] = ch[mx], pos[u] = pos[mx] + 1;
  }
  t[ch[u]].nodes.emplace_back(u);
}
int head(int u) { return t[ch[u]].nodes.back(); }
int ancestor(int u, int v) { return in[u] <= in[v] && in[v] < in[u] + sz[u]; }
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m, q;
  cin >> n >> m >> q;
  for (auto i = (1); i < (n); ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    g[u].emplace_back(v);
    g[v].emplace_back(u);
  }
  for (auto i = (0); i < (m); ++i) {
    int x;
    cin >> x;
    if (int((z[--x]).size()) < K) z[x].emplace_back(i + 1);
  }
  dfs(0, 0);
  for (auto &v : t) v.init();
  while (q--) {
    int u, v;
    cin >> u >> v >> k;
    --u;
    --v;
    vector<int> x;
    while (ch[u] != ch[v]) {
      if (!ancestor(head(u), v)) {
        x = x + t[ch[u]].query(pos[head(u)], pos[u]);
        u = p[head(u)];
      } else {
        x = x + t[ch[v]].query(pos[head(v)], pos[v]);
        v = p[head(v)];
      }
    }
    x = x + t[ch[u]].query(pos[u], pos[v]);
    cout << int((x).size()) << ' ';
    for (auto &z : x) cout << z << ' ';
    cout << '\n';
  }
}